Please help me with this question ASAP at the earliest ???? within an hour will appreciate your help? 2. Lucy has fair (light) skin, red and wavy/curly hair. Use the model of her parents' genes aboveto explain how it's possible for her to have that combination of alleles. Fill out the left part ofthe table, then write your explanation in the right column. In your explanation, make sure touse the words: (underline them)O recessiveO PhenotypeO GenotypeO MomO DadO AlleleO DominantTraitFromFromExplain: Is it possible for Lucy to be her parent's child?dad she Why does she look the way she does? What allele didreceived she get from each parent?themom shereceivedthe allele(one letter alleleper box)(oneletterper box)Hair texture(Hh) CurlySkin color(Dd)FairHair Color(Rr)Red

The process of inheritance involves the transmission of genetic information from parents to progeny.…

2. Lucy has fair (light) skin, red and wavy/curly hair. Use the model of her parents' genes above to explain how it's possible for her to have that combination of alleles. Fill out the left part of the table, then write your explanation in the right column. In your explanation, make sure to use the words: (underline them) O Phenotype O Genotype O Mom O Dad O Allele O Dominant O recessive

2. Lucy has fair (light) skin, red and wavy/curly hair. Use the model of her parents' genes above to explain how it's possible for her to have that combination of alleles. Fill out the left part of the table, then write your explanation in the right column. In your explanation, make sure to use the words: (underline them) O Phenotype O Genotype O Mom O Dad O Allele O Dominant O recessive

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
What type of heredity is shown in the pedigree? (hint: check your notes - "modes of inheritance") 3 Autosomal Dominant Sex-Linked Recessive Autosomal Recessive Sex-Linked Dominant Music off Zoom in 99 esc 26 & 1 2 3 4 5 6 7 8 detete Q w R T. Y P tab A S F K retu caps lock C V B N M shift command option control option command .. ..
! I 7 l 8 / Using the pedigree you have constructed, complete the following Punnett square using A/a. 1) Show the cross between Braxton and his non-albino wife who had an Mother's Alleles albino father: Mother's genotype: Father's genotype: Genotype %: Phenotype %: Click to add speaker notes Father's Alleles
Drag each term to the matching description. (Each box can be used only once. Not all boxes will be used.) 2 dihybrid cross to search * recessive MOSDOV dominant alleles genotype # testcros phenotype 3 * E $ The observable characteristics or traits of an organism. The alleles an individual carries. Alternate forms of a single gene. The allele expressed in a heterozygote's phenotype. 4 The allele does not contribute to a heterozygote's phenotype. dominant A cross between individuals heterozygous for two genes. IOI R O f5 A cross between an ividu of ambiguous genotype with a homozygous recessive individual. Offspring of the F. generation. % !...! 5 31 T fg 1 30 f10 testcross о
Affected wile (whose father was normal) Normal Another couple with a family history of the same disease also come in to see you to obtain genetic counseling. In this case the husband is normal and the wife is affected. The wife's father was not affected by this disease. Determine what their chances are of having a child born with this condition. They would also like to know what the probabilities are of having an affected boy or affected girl. Use the symbols above to complete the diagram right and determine the probabilities stated below (expressed as a proportion or percentage). husband Parents Gametes 5. Determine the probability of having: (a) Affected children: Possible fertilizations (b) An affected girl: O O O O (c) An affected boy: Children Describing examples other than those above, discuss the role of sex linkage in the inheritance of genetic disorders:
G Describe Muller's Ratchet-Googl x nooreps.owschools.com/owsoo/studentAssignment/index?eh=310247513 Asslgnment -6. Mendelian Genetics Attempt 1 of 2 ASSIGNMENTS COURSES SECTION 7 OF 8 « < 4 5 8. 9. 10 11 12 13 14 Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into correct position in the answer box. Release your mouse button when the item is place. If you change your mind, dra the item to the trashcan. Click the trashcan to clear all your answers. Make a Punnett Square for two smooth seed hybrid pea plants (Ss) Click once to select an item at the bottom of the problem. Click again to drop the item in its correct place. S SS Ss SS S
D Page view A Read aloud Draw Highlight PRA Can you curl your tongue up on the sides? Tongue-curling in humans is a dominant genetic trait. Suppose a man who is Tt for tongue-curling marries a woman who is also Tt for this trait. What are the possible genotypes and phenotypes of their children and the percent chance for each? 4. 1 Fill in the table: Genotype Phenotype Chance
Place each box in the appropriate column. (Each box is used only once.) Probability of offspring homozygous recessive for the R gene is 25%. Phenotypic ratio of the progeny will be 1.1. SS Rrx SS rr The chance of Ss Rr offspring is 25%. Probability of an ss Rr offspring is 25%. Probability of an SS offspring is 100%. Phenotypic ratio of the progeny will be 9.3.3:1. Ss RR x Ss rr Genotypic ratio of the progeny will be 1:2:1. Probability of an SS Rr offspring is 50% Ss Rrx Ss Rr
Part B 1. For the remaining characteristics, you will be looking at homozygous and heterozygous traits. Each trait will be determined to be homozygous recessive, homozygous dominant, or hetero- zygous. Letters representing the alleles will be given to you; the uppercase letter is always dominant and the lowercase letter is always recessive. If both dice rolled are even numbers, it indicates homozygous dominant alleles; if both are odd, it indicates homozygous recessive alleles; if one is odd and the other even, it indicates heterozygous alleles. 2. Eyebrows: Use the alleles "A" and "a" for this trait. Two separate brows are the dominant trait, whereas the unibrow (or one continuous brow) is recessive. 3. Eye shape: Use the alleles "R" and "r" for this trait. Round eyes are the dominant trait; almond shaped eyes are recessive. 4. Hitchhiker's thumb: Use the alleles "H" and "h" for this trait. Straight thumbs are dominant; Hitchhiker's thumbs (the curving upward of the thumb) are…
YOUR SISTER DIED FROM TAY-SACHS DISEASE, INHERITED AS A RECESSIVE ALLELE (t). you're married and planning to start your family. you're worried about the disease and decide to have genetic testing to see if you or your spouse is a carrier of the tay-sachs allele. the test results show that you're a carrier of the allele, but your spouse isn't. what is the probability that you and your spouse will have a child with tay-sachs disease? show your work.
This is a schematic representation for a bi-allelic gene with a recessive allele a that causes a specific disease. Use your cursor to select the part of this diagram that represents the homozygous recessive event. Saa AA Aa